SOLUTION: Find the vertex and intercepts for the parabola. Sketch the graph. g(x)=x^2+x-6

Algebra ->  Graphs -> SOLUTION: Find the vertex and intercepts for the parabola. Sketch the graph. g(x)=x^2+x-6      Log On


   

Question 78139: Find the vertex and intercepts for the parabola. Sketch the graph.
g(x)=x^2+x-6
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the vertex, lets complete the square and put the equation in vertex form:
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2%2B1+x-6 Start with the given equation



y%2B6=1+x%5E2%2B1+x Add 6 to both sides



y%2B6=1%28x%5E2%2B1x%29 Factor out the leading coefficient 1



Take half of the x coefficient 1 to get 1%2F2 (ie %281%2F2%29%281%29=1%2F2).


Now square 1%2F2 to get 1%2F4 (ie %281%2F2%29%5E2=%281%2F2%29%281%2F2%29=1%2F4)





y%2B6=1%28x%5E2%2B1x%2B1%2F4-1%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F4 does not change the equation




y%2B6=1%28%28x%2B1%2F2%29%5E2-1%2F4%29 Now factor x%5E2%2B1x%2B1%2F4 to get %28x%2B1%2F2%29%5E2



y%2B6=1%28x%2B1%2F2%29%5E2-1%281%2F4%29 Distribute



y%2B6=1%28x%2B1%2F2%29%5E2-1%2F4 Multiply



y=1%28x%2B1%2F2%29%5E2-1%2F4-6 Now add %2B6 to both sides to isolate y



y=1%28x%2B1%2F2%29%5E2-25%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-1%2F2, and k=-25%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B1x-6 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B1x-6%29 Graph of y=1x%5E2%2B1x-6. Notice how the vertex is (-1%2F2,-25%2F4).



Notice if we graph the final equation y=1%28x%2B1%2F2%29%5E2-25%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B1%2F2%29%5E2-25%2F4%29 Graph of y=1%28x%2B1%2F2%29%5E2-25%2F4. Notice how the vertex is also (-1%2F2,-25%2F4).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.





So the vertex is:
(-0.5, -6.25)


Now lets find the intercepts. The easiest way is to use the quadratic formula:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-6+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+2
x%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B1x%2B-6 can be factored:
1x%5E2%2B1x%2B-6+=+1%28x-2%29%2A%28x--3%29
Again, the answer is: 2, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-6+%29


So the x-intercepts are x=2, x=-3
And here's the graph
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+x%5E2%2Bx-6%29+ graph of y=x%5E2%2Bx-6